POJ 3140 Contestants Division (树dp)

题目链接：http://poj.org/problem?id=3140

题意：

给你一棵树，问你删去一条边，形成的两棵子树的节点权值之差最小是多少。

思路：

dfs

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <vector>
 using namespace std;
 const int N = 1e5 + ;
 typedef long long LL;
 int n, cnt, head[N];
 LL a[N];
 struct Edge {
     int next, to;
 }edge[N << ];
 LL dp[N];

 inline void add(int u, int v) {
     edge[cnt].next = head[u];
     edge[cnt].to = v;
     head[u] = cnt++;
 }

 LL dfs(int u, int p) {
     dp[u] = a[u];
     for(int i = head[u]; ~i; i = edge[i].next) {
         int v = edge[i].to;
         if(v == p)
             continue;
         dp[u] += dfs(v, u);
     }
     return dp[u];
 }

 LL f(LL a) {
     return a >  ? a : -a;
 }

 int main()
 {
     int m, ca = ;
     while(~scanf("%d %d", &n, &m) && (n || m)) {
         LL sum = ;
         for(int i = ; i <= n; ++i) {
             scanf("%lld", a + i);
             sum += a[i];
             head[i] = -;
         }
         cnt = ;
         int u, v;
         while(m--) {
             scanf("%d %d", &u, &v);
             add(u, v);
             add(v, u);
         }
         dfs(, -);
         LL res = sum;
         for(int i = ; i <= n; ++i) {
             res = min(res, f(sum - *dp[i]));
         }
         printf("Case %d: %lld\n", ca++, res);
     }
     return ;
 }